Answer /-
A.0.0931 mol
from ideal gas equation ,
We have ,
PV = nRT
Where P stands where pressure of gas.
We know that pressure of gas at STP = 1 atm
V stands for volume of gas , given Volume is 2.085 L
R stands for gas constant , R = 0.0821
And T stands for absolute temperature and measured in kelvin,T=273K
Now ,
PV = nRT
n = PV/RT
Refer attachment for working
Answer:
6.142 moles of NaCl
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2AlCl3 + 3Na2S —> Al2S3 + 6NaCl
Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:
Mass mass of Na2S = 78.048g/mol
Mass of Na2S = 239.7g
Number of mole Na2S =..?
Mole = Mass /Molar Mass
Number of mole Na2S = 239.7/78.048 = 3.071 moles
Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:
From the balanced equation above,
3 moles of Na2S reacted to produce 6 moles of NaCl.
Therefore, 3.071 moles of Na2S will react to produce = (3.071 x 6)/3 = 6.142 moles of NaCl
<h2>Answer:</h2>
He is right that the energy of vaporization of 47 g of water s 106222 j.
<h3>Explanation:</h3>
Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.
In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.
It means for vaporizing 18 g, 40.65 kJ energy is needed.
So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ
Hence the student is right about the energy of vaporization of 47 g of water.
It is A. Barium
Explanation: I did that already
